Question

A box contains 27 parts, of which 5 are defective and 22 are nondefective. 2 parts are selected without replacement. Not sure how to find the probability that one is defected, both are defective, and neither are defective

Answer 1

so you have 27 parts, probability you get something is

number of things that are like that/27
so for defective things its 5/27, for nondefective its 22/27

so for one being defective, you have two chances to select a defective one so
5/27 + 5/27 = 10/27

for both being defective, = (5/27) x (5/27)
and for non defective = (22/27) x (22/27)

I dont have an intuative reason as to why we multiply related events, hopefully someone here will explain that

Answer 2

thanks you so much!!

Answer 3

however it says that the probability of one being defective is wrong. it needs to be a decimal, so i got .37, which apparently is wrong

Answer 4

Hmm well it says here that you are picking without replacement... meaning that once you pick out a part, the probability of picking a 2nd part effectively changes because there is one less in your box.

Both defective = P(1st is defective AND 2nd is defective) = 5/27 * 4/26
5/27 is the probability of getting 1 defective. Now, since you have already taken 1 part out from the sample, there are 26 parts left in total, and 5-1=4 defective parts left. That's why the second fraction is (4/26)

None defective: The same idea applies. You get (22/27)*(21/26)
22/27 is the probability of getting one non-defective. Now, since you have already taken 1 part out from the sample, there are 26 parts left in total, and 22-1=21 non-defective parts left. That's why the second fraction is (21/26)

One part is defective: This is is a bit trickier.

You can think of why multiplying probabilities for both events occurring makes sense: because the probability of both of them occurring together is smaller than the single individual events (because you are multiplying numbers that are between 0 and 1), and the probability of it occurring both together is a rarer event than the single event. On a smaller scale, say you are flipping a coin. You flip it 2 times. You want to know the probability that two heads occur. This is going to be (1/2)*(1/2) = 1/4, since you can get heads on the first flip, and heads on the 2nd flip. If you list out all the possible outcomes, you see that it agrees with the multiplication rule:
{T, T}
{H, T}
{T, H}
{H, H}

Only the event {H, H} is relevant. And there is 1 of these outcomes out of the 4 total, so the probability is 1/4.

Now maybe you are asking why I did an addition mixed with the multiplication in the "1 defective part" case:
If you have different CASES in which an event can occur.. like you could have, in your example
-1st box part is defective AND 2nd box is non-defective (A)

-OR: 1st box part is non-defective AND 2nd box is defective (B)

Then (A) is : (5/27)(22/26) = 55/351
And (B) is :(22/27)(5/26) = 55/351

Now you add both cases: 55/351 + 55/351

We have these 2 separate cases since we don't know if one was pulled from the box before the other, or vice versa, so you need to think of all those probabilities. Back to the coin example: if I said what is the probability of getting one heads... you have two cases since you don't know if Tail occurred first, or heads:
{H, T}
{T, H}
And you would add those cases together.

Because you have 2 separate cases (notice the "OR" ) that can occur (and they CANNOT occur together), then you must add those probabilities. The "AND" statement I wrote because it represents the multiplication of the outcome occurring simultaneously of 1 defective with 1 non-defective

Back to the coin example: If I asked "What is the probability of getting two heads, or no heads?" Then, you would have the events
{T, T} , {H, H}. So you'd get (1/4)+(1/4) = 1/2

One trick: If you can try re-phrasing your questions into a combination of "AND" and "OR" statements, you can more quickly see when to use the multiplication and addition.
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An alternative to all this... if you are comfortable with combinations (n choose r), then this whole problem can be done by using those. (Maybe google "hypergeometric distribution")

Answer 5

what does the drawing have to do with the problem!!!!!
for the one being defective you've got two chances
5/27+5/27=10/27